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import warnings
warnings.filterwarnings('ignore')
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import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
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from sklearn.cluster import KMeans
from sklearn.cluster import AgglomerativeClustering
from sklearn import metrics
from scipy.stats import zscore
from sklearn.preprocessing import StandardScaler
from scipy.spatial.distance import cdist
from scipy.spatial.distance import pdist
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from scipy.cluster.hierarchy import cophenet, dendrogram, linkage
from scipy.cluster.hierarchy import fcluster
from sklearn.metrics import silhouette_score
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df = pd.read_excel("Credit Card Customer Data.xlsx")
df.head()
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#1
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df.shape
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#Shape, Data Types
df.info()
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#Removing 'Sl_No' column as it all the unique value and this column wont provide any insight to build a model
df.drop('Sl_No', axis=1, inplace=True)
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df.describe().transpose()
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df.nunique()
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# There are 655 Unique Customer key. So, there must be 5 duplicate customer keys
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#Check All duplicate Customer Keys.
df[df.duplicated(subset=['Customer Key'],keep=False)]
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#5 duplicate customer keys should be removed
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for i in df.columns:
sns.distplot(df[i],hist=False,)
plt.show()
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#Looks like there are 3 or 4 clusters. Need to use elbow method to confirm
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#Check for outliers using box plots
plt.figure(figsize=(15,10))
pos = 1
for i in df.columns:
plt.subplot(3, 3, pos)
sns.boxplot(df[i])
pos += 1
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# There are several outliers for Average Credit limit and Total Visits Online
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corr = df.corr()
sns.heatmap(corr, annot = True)
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#Total_calls_made is highly negatively correlated with Total_Credit_Cards
#Total_calls_made is negatively correlated with Avg_Credit_Limit and Total_visits_bank
#Total_visits_online is negatively correlated with Total_visits_bank
#Avg_Credit_Limit is positively correlated with Total_Credit_Cards and Total_visits_online
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sns.pairplot(df,diag_kind='kde');
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#Looks like we might need 4 clusters based on Total_Credit_Cards. Let's try elbow method to confirm
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import pandas_profiling
#Getting the pandas profiling report and check for incorrect imputation
pandas_profiling.ProfileReport(df)
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#2
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#There are duplicate Customer key. We need to remove the duplicates
#Based on the graphs we might need 3 or 4 clusters, we can confirm the same using the elbow method later.
#There are several outliers for Avg_Credit_Limit and Total_visits_online
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df[df.duplicated(subset=['Customer Key'],keep='last')]
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df.drop_duplicates(subset=['Customer Key'],keep='last', inplace=True)
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#To confirm duplicates are dropped
df[df.duplicated(subset=['Customer Key'],keep='last')]
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##Scale the data using zscore
df_z = df.apply(zscore)
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#3
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#### Finding optimal no. of clusters for scaled data
clusters=range(1,10)
meanDistortions1=[]
for k in clusters:
model1=KMeans(n_clusters=k)
model1.fit(df_z)
prediction=model1.predict(df_z)
meanDistortions1.append(sum(np.min(cdist(df_z, model1.cluster_centers_, 'euclidean'), axis=1)) / df_z
.shape[0])
plt.plot(clusters, meanDistortions1, 'bx-')
plt.xlabel('k')
plt.ylabel('Average distortion')
plt.title('Selecting k with the Elbow Method')
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#Based on Elbow Method, 3 or 5 clusters.
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#Let's try the value of 3 for K first
kmeans3 = KMeans(n_clusters=3, n_init = 15, random_state=1)
kmeans3.fit(df_z)
k3_prediction=kmeans3.predict(df_z)
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#Append the prediction
df["K3_GROUP"] = k3_prediction
df_z["K3_GROUP"] = k3_prediction
df.head()
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df_z.head(10)
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#Analyze the distribution of the data among the three groups (K = 3)
df_groupedk3 = df_z.groupby(['K3_GROUP'])
df_groupedk3.mean()
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#Perform Value counts
df_groupedk3['K3_GROUP'].value_counts()
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df_z.boxplot(by = 'K3_GROUP', layout=(2,4), figsize=(20, 15))
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#Let's try the value of 5 for K as the next elbow is at 5
kmeans5 = KMeans(n_clusters=5, n_init = 15, random_state=1)
kmeans5.fit(df_z)
k5_prediction=kmeans5.predict(df_z)
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#Append the k-means prediction for K=5
df["K5_GROUP"] = k5_prediction
df_z["K5_GROUP"] = k5_prediction
df_z.head(10)
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#Analyze the distribution of the data among the three groups (K = 3)
df_groupedk5 = df_z.groupby(['K5_GROUP'])
df_groupedk5.mean()
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#Perform Value counts for K=5
df_groupedk5['K5_GROUP'].value_counts()
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#Box plots for K=5
df_z.boxplot(by = 'K5_GROUP', layout=(2,4), figsize=(40, 30))
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#4
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# cophenet index is a measure of the correlation between the distance of points in feature space and distance on dendrogram
# closer it is to 1, the better is the clustering
Z_ea = linkage(df_z, metric='euclidean', method='average')
c_ea, coph_dists_ea = cophenet(Z_ea , pdist(df_z))
print("cophenet index for Average method with Euclidean distance :",c_ea)
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plt.figure(figsize=(10, 5))
plt.title('Agglomerative Hierarchical Clustering Dendogram')
plt.xlabel('sample index')
plt.ylabel('Distance')
dendrogram(Z_ea, leaf_rotation=90.,color_threshold = 40, leaf_font_size=8. )
plt.tight_layout()
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Z_ec = linkage(df_z, metric='euclidean', method='complete')
c_ec, coph_dists_ec = cophenet(Z_ec , pdist(df_z))
print("cophenet index for Complete method with Euclidean distance :",c_ec)
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plt.figure(figsize=(10, 5))
plt.title('Agglomerative Hierarchical Clustering Dendogram')
plt.xlabel('sample index')
plt.ylabel('Distance')
dendrogram(Z_ec, leaf_rotation=90.,color_threshold = 40, leaf_font_size=8. )
plt.tight_layout()
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Z_ew = linkage(df_z, metric='euclidean', method='ward')
c_ew, coph_dists_ew = cophenet(Z_ew , pdist(df_z))
print("cophenet index for Ward method with Euclidean distance :",c_ew)
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plt.figure(figsize=(10, 5))
plt.title('Agglomerative Hierarchical Clustering Dendogram')
plt.xlabel('sample index')
plt.ylabel('Distance')
dendrogram(Z_ew, leaf_rotation=90.,color_threshold = 40, leaf_font_size=8. )
plt.tight_layout()
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# Cophenet index for Euclidean distance is maximum for Average method : 0.9185047026907476
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model_a3 = AgglomerativeClustering(n_clusters=3, affinity='euclidean', linkage='average')
model_a3.fit(df_z)
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#Append the Agglo prediction for 3 clusters
df["A3_GROUP"] = model_a3.labels_
df_z["A3_GROUP"] = model_a3.labels_
df_z.head(10)
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#Analyze the distribution of the data among the three CLUSTERS
df_groupeda3 = df_z.groupby(['A3_GROUP'])
df_groupeda3.mean()
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df_z['A3_GROUP'].value_counts()
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#Box plots for 3 clusters
df_z.boxplot(by = 'A3_GROUP', layout=(4,8), figsize=(40, 30))
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#5
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#Silhouette Score for Hierarchical clustering
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list_clusters = [2, 3, 4, 5, 6]
Total_score_loop = 0
for i in list_clusters:
model_loop = AgglomerativeClustering(n_clusters=i, affinity='euclidean', linkage='average')
model_loop.fit(df_z)
score_loop = silhouette_score(df_z, model_loop.fit_predict(df_z))
Total_score_loop = Total_score_loop + score_loop
print('Silhouette Score for Agglo with', i ,'clusters: %.2f' % score_loop)
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print('Total Silhouette Score for AgglomerativeClustering : %.2f' % Total_score_loop)
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Avg_score = Total_score_loop/5
print('Average Silhouette Score for AgglomerativeClustering : %.2f' % Avg_score)
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#Silhouette Score for K-means clustering
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list_clusters = [2, 3, 4, 5, 6]
Total_score_kloop = 0
for i in list_clusters:
kmeans_loop = KMeans(n_clusters=i, n_init = 15, random_state=1)
kmeans_loop.fit(df_z)
kmeans_loop_prediction=kmeans_loop.predict(df_z)
score_kloop = silhouette_score(df_z, kmeans_loop.labels_, metric='euclidean')
Total_score_kloop = Total_score_kloop + score_kloop
print('Silhouette Score for k-Means with', i ,'clusters: %.2f' % score_kloop)
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print('Total Silhouette Score for k-Means Clustering : %.2f' % Total_score_kloop)
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Avg_kscore = Total_score_kloop/5
print('Average Silhouette Score for k-Means Clustering : %.2f' % Avg_kscore)
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#The silhouette ranges from −1 to +1, where a high value indicates that the object is well matched to its own cluster and poorly matched to neighboring clusters.
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#Based on Silhouette score 4 clusters are good for both K-Means and Hierarchical.
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#Using K=4 as it has the highest Silhouette score
kmeans4 = KMeans(n_clusters=4, n_init = 14, random_state=1)
kmeans4.fit(df_z)
k4_prediction=kmeans4.predict(df_z)
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#Append the k-means prediction for K=4
df["K4_GROUP"] = k4_prediction
df_z["K4_GROUP"] = k4_prediction
df_z.head(10)
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k_centroids = kmeans4.cluster_centers_
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k_centroids
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#Analyze the distribution of the data among the four groups (K = 4)
df_groupedk4 = df_z.groupby(['K4_GROUP'])
df_groupedk4.mean()
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#Perform Value counts for K=4
df_groupedk4['K4_GROUP'].value_counts()
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#Box plots for K=4
df_z.boxplot(by = 'K4_GROUP', layout=(3,5), figsize=(40, 30))
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from mpl_toolkits.mplot3d import Axes3D
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## 3D plots of clusters
fig = plt.figure(figsize=(8, 6))
ax = Axes3D(fig, rect=[0, 0, .95, 1], elev=20, azim=60)
labels = kmeans4.labels_
ax.scatter(df_z.iloc[:, 0], df_z.iloc[:, 1], df_z.iloc[:, 2],c=labels.astype(np.float), edgecolor='k')
ax.w_xaxis.set_ticklabels([])
ax.w_yaxis.set_ticklabels([])
ax.w_zaxis.set_ticklabels([])
ax.set_xlabel('Length')
ax.set_ylabel('Height')
ax.set_zlabel('Weight')
ax.set_title('3D plot of KMeans Clustering')
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#AgglomerativeClustering for 4 clusters
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model_a4 = AgglomerativeClustering(n_clusters=4, affinity='euclidean', linkage='average')
model_a4.fit(df_z)
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#Append the Agglo prediction for 4 clusters
df["A4_GROUP"] = model_a4.labels_
df_z["A4_GROUP"] = model_a4.labels_
df_z.head(10)
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#Analyze the distribution of the data among the four CLUSTERS
df_groupeda4 = df_z.groupby(['A4_GROUP'])
df_groupeda4.mean()
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df_z['A4_GROUP'].value_counts()
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#Box plots for 4 clusters
df_z.boxplot(by = 'A4_GROUP', layout=(4,8), figsize=(40, 40))
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dendrogram(
Z_ea,
truncate_mode='lastp', # show only the last p merged clusters
p=4, # show only the last p merged clusters
)
plt.show()
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#6
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#Compare 4 cluster for K-Means and AgglomerativeClustering
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df_z['A4_GROUP'].value_counts()
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df_z['K4_GROUP'].value_counts()
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##Based of the value counts,
#Cluster 0 in AgglomerativeClustering is same as Cluster 2 in K-Means clustering
#Cluster 2 in AgglomerativeClustering is same as Cluster 1 in K-Means clustering
#Cluster 3 in AgglomerativeClustering is same as Cluster 0 in K-Means clustering
#Cluster 1 in AgglomerativeClustering is same as Cluster 3 in K-Means clustering
#Mean below also confirms that
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df_groupeda4.mean()
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#7
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##Group 0 is customers with medium Avg_Credit_Limit, medium Total_Credit_Cards,
#highest Total_visits_bank, low Total_visits_online and low Total_calls_made
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##Group 1 is customers with highest Avg_Credit_Limit, highest Total_Credit_Cards,
#least Total_visits_bank, highest Total_visits_online and least Total_calls_made
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##Group 2 is customers with least Avg_Credit_Limit, least Total_Credit_Cards,
#low Total_visits_bank, medium Total_visits_online and highest Total_calls_made
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##Group 3 is customers with low Avg_Credit_Limit, low Total_Credit_Cards,
#low Total_visits_bank, medium Total_visits_online and medium Total_calls_made
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#Key Questions:
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##How many different segments of customers are there?
#There are 4 different segments of customers
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##How are these segments different from each other?
#As already mentioned above
##Group 0 is customers with medium Avg_Credit_Limit, medium Total_Credit_Cards,
#high Total_visits_bank, low Total_visits_online and low Total_calls_made
##Group 1 is customers with highest Avg_Credit_Limit, highest Total_Credit_Cards,
#least Total_visits_bank, highest Total_visits_online and least Total_calls_made
##Group 2 is customers with least Avg_Credit_Limit, least Total_Credit_Cards,
#low Total_visits_bank, medium Total_visits_online and highest Total_calls_made
##Group 3 is customers with low Avg_Credit_Limit, low Total_Credit_Cards,
#low Total_visits_bank, medium Total_visits_online and medium Total_calls_made
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##What are your recommendations to the bank on how to better market to and service these customers?
#Number of customers are high is Group 0 i.e., customers with medium Avg_Credit_Limit,
#medium Total_Credit_Cards, high Total_visits_bank, low Total_visits_online and low Total_calls_made
#Bank can concentrate more of these customers as they are high in number
##Customers with highest Total_Credit_Cards are using more online banking, so online marketing
#will work best for these customers
##Customers with least Avg_Credit_Limit and least Total_Credit_Cards are making more
#calls to the bank, so these customer can be reached by phone or phone marketing
#will work best for these customers
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